A `10 muC` charge is divided into two parts and placed at `1 cm` distance so that the repusive force between them is maximum . The charges of the two parts are :

Options

A. `9 muC, 1 muC`
B. `7 muC, 3 muC`
C. `8 muC, 2 muC`
D. `5 muC, 5 muC`

Answer

To solve the problem of dividing a `10 µC` charge into two parts such that the repulsive force between them is maximized, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**: We have a total charge \( Q = 10 \, \mu C \) that is divided into two parts, \( q_1 \) and \( q_2 \), such that \( q_1 + q_2 = Q \). The distance between the two charges is \( r = 1 \, cm = 0.01 \, m \).

2. **Setting Up the Force Equation**: The electrostatic force \( F \) between two point charges is given by Coulomb's law:
\[ F = k \frac{q_1 q_2}{r^2} \]
where \( k \) is Coulomb's constant. Since \( q_2 = Q - q_1 \), we can rewrite the force as:
\[ F = k \frac{q_1 (Q - q_1)}{r^2} \]

3. **Substituting Known Values**: Substitute \( Q = 10 \, \mu C \) and \( r = 0.01 \, m \):
\[ F = k \frac{q_1 (10 \, \mu C - q_1)}{(0.01)^2} \]

4. **Maximizing the Force**: To find the maximum force, we need to differentiate \( F \) with respect to \( q_1 \) and set the derivative to zero:
\[ F = k \frac{(10 \, \mu C) q_1 - q_1^2}{(0.01)^2} \]
Differentiate:
\[ \frac{dF}{dq_1} = k \frac{(10 \, \mu C) - 2q_1}{(0.01)^2} \]
Set the derivative to zero:
\[ 10 \, \mu C - 2q_1 = 0 \]
This gives:
\[ q_1 = 5 \, \mu C \]

5. **Finding the Other Charge**: Since \( q_2 = Q - q_1 \):
\[ q_2 = 10 \, \mu C - 5 \, \mu C = 5 \, \mu C \]

6. **Conclusion**: The two charges that maximize the repulsive force between them when the total charge is \( 10 \, \mu C \) and they are placed \( 1 \, cm \) apart are:
\[ q_1 = 5 \, \mu C \quad \text{and} \quad q_2 = 5 \, \mu C \]

### Final Answer:
The charges of the two parts are \( 5 \, \mu C \) and \( 5 \, \mu C \).

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